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$$\sum_{n=1}^\infty\frac{\cot(\pi n\sqrt{61})}{n^3}=-\frac{16793\pi^3}{45660\sqrt{61}}.$$

  • Prove it converges and,

  • evaluate the series.


For the first part of the question, I prove it converges by considering the irrationality measure,

$$|\sin (n\pi\sqrt{61})|=|\sin (n\pi\sqrt{61}-m\pi)|\ge \frac{2}{\pi}|n\pi\sqrt{61}-m\pi|\ge 2n\left|\sqrt{61}-\frac{m}{n}\right|>\frac{C}{n},$$

so $$\left|\frac{\cot(\pi n\sqrt{61})}{n^3}\right|\le\left|\frac{1}{n^3\sin(\pi n\sqrt{61})}\right|<\frac{C}{n^2}.$$

How to evaluate the series?

I have found a related question.

I apologize for incorrect information in the previous post.

Tianlalu
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2 Answers2

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In general $$F(\alpha) = \sum_{n=1}^\infty \frac{\cot(n\pi \alpha)}{n^3}$$ converges and can be explicitly calculated when $\alpha$ is a quadratic irrational. The convergence in this case is easily seen as $\alpha$ has irrationality measure $2$. More precisely, $F(\alpha)/\pi^3 \in \mathbb{Q}(\alpha)$ when $\alpha$ is quadratic irrational.

The procedure below also works when $n^3$ is replaced by any $n^{2k+1}$.


Let $$g(z) = \frac{\cot(z\pi \alpha)\cot(z\pi)}{z^3}$$ then $g$ has simple poles at non-zero integer multiples of $1$ and $1/\alpha$, and $5$-th order pole at $0$. Let $R_N$ denote a large rectangle with corners at $N(\pm 1 \pm i)$. Then contour integration gives $$\tag{1}\sum_{\substack{n\in R_N \\ n\neq 0}} \frac{\cot(n\pi\alpha)}{\pi n^3} + \sum_{\substack{n/\alpha\in R_N \\ n\neq 0}} \frac{\alpha^2\cot(n\pi/\alpha)}{\pi n^3}-\frac{\pi ^2 \left(\alpha^4-5 \alpha^2+1\right)}{45 \alpha} = \frac{1}{2\pi i} \int_{R_N} g(z)dz$$ I claim there exists a sequence of integers $N_1, N_2, \cdots$ such that RHS tends to $0$. Note that $\cot(z\pi)$ is uniformly bounded on the annulus $R_{N+3/4} - R_{N+1/4}$ when $N$ is an integer. Hence by equidistribution of $n\alpha$ modulo $1$, we can find integers $N_i$ such that both $\cot(z\pi\alpha)$ and $\cot(z\pi)$ are uniformly bounded on $R_{N_i+3/4} - R_{N_i+1/4}$.

Since we already know the series converges, from $(1)$: $$\tag{2}F(\alpha) + \alpha^2F(\frac{1}{\alpha}) = \underbrace{\frac{\pi ^3 \left(\alpha^4-5 \alpha^2+1\right)}{90 \alpha}}_{\rho(\alpha)}$$ Note that obviously $F(\alpha+1)=F(\alpha)$.


Let the continued fraction expansion of $\alpha$ be given by $$\alpha = [a_0;a_1,a_2,\cdots]$$ Successive complete quotients are denoted by: $$\zeta_0 = [a_0;a_1,a_2,\cdots]\qquad \zeta_1 = [a_1;a_2,a_3,\cdots]\qquad \zeta_2 = [a_2;a_3,a_4,\cdots]$$ Then $(2)$ and periodicity implies for $k\geq 0$: $$\tag{3} F(\zeta_{k+1}) + \zeta_{k+1}^2 F(\zeta_k) = \rho(\zeta_{k+1})$$ If continued fraction of $\alpha$ is of form $$\alpha = [a_0;a_1,\cdots,a_m,\overline{b_1,\cdots,b_r}]$$ Then $\zeta_{m+r+1} = \zeta_{m+1}$, so we eventually entered a cycle. $(3)$ gives a system of $m+r+1$ linear equations (by setting $k=0,\cdots,m+r$), with $m+r+1$ variables: $F(\zeta_0), F(\zeta_1),\cdots,F(\zeta_{m+r})$. $$\begin{cases} F(\zeta_1) + \zeta_1^2 F(\zeta_0) &= \rho(\zeta_1) \\ F(\zeta_2) + \zeta_2^2 F(\zeta_1) &= \rho(\zeta_2) \\ \cdots \\ F(\zeta_{m+1}) + \zeta_{m+1}^2 F(\zeta_{m+r}) &= \rho(\zeta_{m+1}) \end{cases}$$ Solving it gives the value of $F(\zeta_0)=F(\alpha)$.


For $\alpha = \sqrt{61} = [7;\overline{1,4,3,1,2,2,1,3,4,1,14}]$, we have $$\begin{aligned} \zeta_0 = \sqrt{61} \qquad \zeta_1 &= \frac{1}{12}(7+\sqrt{61}) \\ \zeta_2 = \frac{1}{3}(5+\sqrt{61}) \qquad \zeta_3 &= \frac{1}{4}(7+\sqrt{61})\\ \zeta_4 = \frac{1}{9}(5+\sqrt{61}) \qquad \zeta_5 &= \frac{1}{5}(4+\sqrt{61})\\ \zeta_6 = \frac{1}{5}(6+\sqrt{61}) \qquad \zeta_7 &= \frac{1}{9}(4+\sqrt{61})\\ \zeta_8 = \frac{1}{4}(5+\sqrt{61}) \qquad \zeta_9 &= \frac{1}{3}(7+\sqrt{61}) \\ \zeta_{10} = \frac{1}{12}(5+\sqrt{61}) \qquad \zeta_{11} &= \frac{1}{12}(7+\sqrt{61}) \end{aligned}$$ solving the above system gives the result.


A few examples: for $\alpha = (1+\sqrt{5})/2$, the continued fraction has period $1$, direct substitution into $(2)$ gives $$\sum_{n=1}^\infty \frac{\cot(n\pi \frac{1+\sqrt{5}}{2})}{n^3} = -\frac{\pi ^3}{45 \sqrt{5}}$$ Complexity of result increases as period of $\alpha$ increases. For $\alpha = \sqrt{211}$, which has period $26$: $$\sum_{n=1}^\infty \frac{\cot(n\pi \sqrt{211})}{n^3} = \frac{128833758679 \pi ^3}{383254107060 \sqrt{211}}$$ For $\alpha = \sqrt{1051}$, with period $50$: $$\sum_{n=1}^\infty \frac{\cot(n\pi \sqrt{1051})}{n^3} = \frac{47332791433774124737806821 \pi ^3}{589394448213331173141730140 \sqrt{1051}}$$ When $\alpha$ is not a "pure" quadratic irrational, the result involves "constant term" (because of non-trivial automorphism of $\mathbb{Q}(\alpha)$): $$\sum_{n=1}^\infty \frac{\cot(n\pi(\frac{1}{4}+\frac{\sqrt{7}}{3}))}{n^3} = \frac{13 \pi ^3}{288}+\frac{104771 \pi ^3}{1244160 \sqrt{7}}$$ The closed-form here follows immediately by noting $\csc x = \cot (x/2) - \cot x$.


I wrote a Mathematica code to evaluate this sum. The command cotsum[Sqrt[61]] evaluates the sum in the question. You can try other quadratic irrationals as well.

This algorithm can be made more efficient, but I don't have much motivation to optimize it.

cotsum[x_] /; QuadraticIrrationalQ[x] := 
 Module[{a1 = x, list, l, r, i, nlist, solution, output, equation, 
   string}, list = ContinuedFraction[a1];
  l = Length[list] - 1;
  r = Length[list[[l + 1]]]; Global`f[a_] := (1 - 5 a^2 + a^4)/90/a;
  i = 1; string = "{";
  While[i < l + r + 1, string = string <> "x" <> ToString[i] <> ",";
   i++]; string = StringTake[string, StringLength[string] - 1] <> "}";
  Do[Evaluate[ToExpression["a" <> ToString[i + 1]]] = 
    FromContinuedFraction[Drop[list, i]], {i, 1, l - 1}];
  nlist = list[[l + 1]];
  Do[Evaluate[ToExpression["a" <> ToString[i + l + 1]]] = 
    FromContinuedFraction[{Flatten[
       Append[Drop[nlist, i], Take[nlist, i]]]}], {i, 0, r - 1}];
  equation = 
   Table[ToExpression[
     "x" <> ToString[i + 1] <> "+a" <> ToString[i + 1] <> "^2*x" <> 
      ToString[i] <> "==f[a" <> ToString[i + 1] <> "]"], {i, 1, 
     r + l - 1}];
  equation = 
   Append[equation, 
    ToExpression[
     "x" <> ToString[l + 1] <> "+a" <> ToString[l + 1] <> "^2*x" <> 
      ToString[r + l] <> "==f[a" <> ToString[l + 1] <> "]"]];
  solution = Solve[equation, ToExpression[string]]; Clear["a*"];
  output = (ToExpression[string][[1]] /. Flatten[solution])*Pi^3; 
  Clear[f];
  FullSimplify[output]]
pisco
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  • This is amazing. – Szeto Nov 30 '18 at 10:44
  • (+1) Very good explanation. – user90369 Nov 30 '18 at 10:49
  • Truly perfect and awesome.. Thank you very much! – Tianlalu Nov 30 '18 at 11:28
  • It seems your notation with the $\zeta_j$ gives exactly the middle column in Prof. Lubin's classroom method for solving Pell's equation. I posted that as an answer, I have it written as a program that produces the Latex code in full, ready for pasting in an answer box on this site. I suspect Prof. Lubin's method is very similar to what Fermat, Wallis, and Brouncker would have used, being entirely made up of integer operations. – Will Jagy Nov 30 '18 at 18:40
  • (+1) Well done. – Mark Viola Nov 30 '18 at 19:41
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    +1 and it appears Ramanujan also said the same thing. See https://math.stackexchange.com/q/907480/72031 – Paramanand Singh Dec 01 '18 at 18:10
  • @ParamanandSingh Yes, this is the case when $\alpha$ is purely imaginary in $(2)$. I think the sum $\sum \coth(\pi n)/n^3$ is also due to Ramanujan. – pisco Dec 02 '18 at 06:33
  • I think there is a general formula for such sums which Ramanujan proved. Here is one particular example https://math.stackexchange.com/q/991066/72031 – Paramanand Singh Dec 02 '18 at 13:15
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Sigh. I believe that Pisco's sequence from $\zeta_1$ to $\zeta_{11}$ matches the middle column (all plus signs) in Prof. Lubin's no-calculator method for solving Pell, and is likely what Fermat, Wallis, Brouncker used all those years ago. Yes! If I continued one more step, I would get $\zeta_{11},$ which equals $\zeta_1$

Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$

$$ \sqrt { 61} = 7 + \frac{ \sqrt {61} - 7 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {61} - 7 } = \frac{ \sqrt {61} + 7 }{12 } = 1 + \frac{ \sqrt {61} - 5 }{12 } $$ $$ \frac{ 12 }{ \sqrt {61} - 5 } = \frac{ \sqrt {61} + 5 }{3 } = 4 + \frac{ \sqrt {61} - 7 }{3 } $$ $$ \frac{ 3 }{ \sqrt {61} - 7 } = \frac{ \sqrt {61} + 7 }{4 } = 3 + \frac{ \sqrt {61} - 5 }{4 } $$ $$ \frac{ 4 }{ \sqrt {61} - 5 } = \frac{ \sqrt {61} + 5 }{9 } = 1 + \frac{ \sqrt {61} - 4 }{9 } $$ $$ \frac{ 9 }{ \sqrt {61} - 4 } = \frac{ \sqrt {61} + 4 }{5 } = 2 + \frac{ \sqrt {61} - 6 }{5 } $$ $$ \frac{ 5 }{ \sqrt {61} - 6 } = \frac{ \sqrt {61} + 6 }{5 } = 2 + \frac{ \sqrt {61} - 4 }{5 } $$ $$ \frac{ 5 }{ \sqrt {61} - 4 } = \frac{ \sqrt {61} + 4 }{9 } = 1 + \frac{ \sqrt {61} - 5 }{9 } $$ $$ \frac{ 9 }{ \sqrt {61} - 5 } = \frac{ \sqrt {61} + 5 }{4 } = 3 + \frac{ \sqrt {61} - 7 }{4 } $$ $$ \frac{ 4 }{ \sqrt {61} - 7 } = \frac{ \sqrt {61} + 7 }{3 } = 4 + \frac{ \sqrt {61} - 5 }{3 } $$ $$ \frac{ 3 }{ \sqrt {61} - 5 } = \frac{ \sqrt {61} + 5 }{12 } = 1 + \frac{ \sqrt {61} - 7 }{12 } $$ $$ \frac{ 12 }{ \sqrt {61} - 7 } = \frac{ \sqrt {61} + 7 }{1 } = 14 + \frac{ \sqrt {61} - 7 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccccccccccccccccccccccccccccccccc} & & 7 & & 1 & & 4 & & 3 & & 1 & & 2 & & 2 & & 1 & & 3 & & 4 & & 1 & & 14 & & 1 & & 4 & & 3 & & 1 & & 2 & & 2 & & 1 & & 3 & & 4 & & 1 & & 14 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 7 }{ 1 } & & \frac{ 8 }{ 1 } & & \frac{ 39 }{ 5 } & & \frac{ 125 }{ 16 } & & \frac{ 164 }{ 21 } & & \frac{ 453 }{ 58 } & & \frac{ 1070 }{ 137 } & & \frac{ 1523 }{ 195 } & & \frac{ 5639 }{ 722 } & & \frac{ 24079 }{ 3083 } & & \frac{ 29718 }{ 3805 } & & \frac{ 440131 }{ 56353 } & & \frac{ 469849 }{ 60158 } & & \frac{ 2319527 }{ 296985 } & & \frac{ 7428430 }{ 951113 } & & \frac{ 9747957 }{ 1248098 } & & \frac{ 26924344 }{ 3447309 } & & \frac{ 63596645 }{ 8142716 } & & \frac{ 90520989 }{ 11590025 } & & \frac{ 335159612 }{ 42912791 } & & \frac{ 1431159437 }{ 183241189 } & & \frac{ 1766319049 }{ 226153980 } \\ \\ & 1 & & -12 & & 3 & & -4 & & 9 & & -5 & & 5 & & -9 & & 4 & & -3 & & 12 & & -1 & & 12 & & -3 & & 4 & & -9 & & 5 & & -5 & & 9 & & -4 & & 3 & & -12 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 61 \cdot 0^2 = 1 & \mbox{digit} & 7 \\ \frac{ 7 }{ 1 } & 7^2 - 61 \cdot 1^2 = -12 & \mbox{digit} & 1 \\ \frac{ 8 }{ 1 } & 8^2 - 61 \cdot 1^2 = 3 & \mbox{digit} & 4 \\ \frac{ 39 }{ 5 } & 39^2 - 61 \cdot 5^2 = -4 & \mbox{digit} & 3 \\ \frac{ 125 }{ 16 } & 125^2 - 61 \cdot 16^2 = 9 & \mbox{digit} & 1 \\ \frac{ 164 }{ 21 } & 164^2 - 61 \cdot 21^2 = -5 & \mbox{digit} & 2 \\ \frac{ 453 }{ 58 } & 453^2 - 61 \cdot 58^2 = 5 & \mbox{digit} & 2 \\ \frac{ 1070 }{ 137 } & 1070^2 - 61 \cdot 137^2 = -9 & \mbox{digit} & 1 \\ \frac{ 1523 }{ 195 } & 1523^2 - 61 \cdot 195^2 = 4 & \mbox{digit} & 3 \\ \frac{ 5639 }{ 722 } & 5639^2 - 61 \cdot 722^2 = -3 & \mbox{digit} & 4 \\ \frac{ 24079 }{ 3083 } & 24079^2 - 61 \cdot 3083^2 = 12 & \mbox{digit} & 1 \\ \frac{ 29718 }{ 3805 } & 29718^2 - 61 \cdot 3805^2 = -1 & \mbox{digit} & 14 \\ \frac{ 440131 }{ 56353 } & 440131^2 - 61 \cdot 56353^2 = 12 & \mbox{digit} & 1 \\ \frac{ 469849 }{ 60158 } & 469849^2 - 61 \cdot 60158^2 = -3 & \mbox{digit} & 4 \\ \frac{ 2319527 }{ 296985 } & 2319527^2 - 61 \cdot 296985^2 = 4 & \mbox{digit} & 3 \\ \frac{ 7428430 }{ 951113 } & 7428430^2 - 61 \cdot 951113^2 = -9 & \mbox{digit} & 1 \\ \frac{ 9747957 }{ 1248098 } & 9747957^2 - 61 \cdot 1248098^2 = 5 & \mbox{digit} & 2 \\ \frac{ 26924344 }{ 3447309 } & 26924344^2 - 61 \cdot 3447309^2 = -5 & \mbox{digit} & 2 \\ \frac{ 63596645 }{ 8142716 } & 63596645^2 - 61 \cdot 8142716^2 = 9 & \mbox{digit} & 1 \\ \frac{ 90520989 }{ 11590025 } & 90520989^2 - 61 \cdot 11590025^2 = -4 & \mbox{digit} & 3 \\ \frac{ 335159612 }{ 42912791 } & 335159612^2 - 61 \cdot 42912791^2 = 3 & \mbox{digit} & 4 \\ \frac{ 1431159437 }{ 183241189 } & 1431159437^2 - 61 \cdot 183241189^2 = -12 & \mbox{digit} & 1 \\ \frac{ 1766319049 }{ 226153980 } & 1766319049^2 - 61 \cdot 226153980^2 = 1 & \mbox{digit} & 14 \\ \end{array} $$

Will Jagy
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